Comments (2)
这儿是不是多写了个负号呀?
d=-x3.dot((x1 - x3).cross(x2-x3)) = -x3.dot(x1.cross(x2))
或者 ?
n = (x1 -x0).cross(x2 - x0);
d = n.transpose() * x0;
//d = n.transpose() * x1;
//d = n.transpose() * x2;
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没有多写负号,我觉得是这样:
n = (x1 - x3).cross(x2 - x3)
d = -x3.dot(n)
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